[Proof of Lemma REF ]
We basically follows the proof of [1]} here.
First, note that
\(\int _{0}^{t} \Big (\frac{u}{\epsilon }\Big )^{\frac{1}{4}} \Big [\log \frac{u}{\epsilon }\Big ]^k \, d\epsilon &= u \int _{\tau }^{+\infty } e^{-\frac{3}{4}v} v^k \, dv, & v = \log \frac{u}{\epsilon }\) 
